STRUCTURE OF COPPER ISOTOPES
.By Prof. Lefteris Kaliambos (Λευτέρης Καλιαμπός)T.E. Institute of Larissa Greece. (August 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ”(2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Copper (Cu) has two stable isotopes, Cu-63 and Cu-65, along with 27 radioisotopes. The most stable of these is Cu-67 with a half-life of 61.83 hours. The least stable is Cu-54 with a half-life of approximately 75 ns. Most have half-lives under a minute. Unstable copper isotopes with atomic masses below 63 tend to undergo β+ decay, while isotopes with atomic masses above 65 tend to undergo β− decay. Cu-64 decays by both β+ and β−. Cu-68, Cu-69, Cu-71, Cu-72, and Cu-76 each have one metastable isomer. Cu-70 has two isomers, making a total of 7 distinct isomers. The most stable of these is Cu-68m with a half-life of 3.75 minutes. The least stable is Cu-69m with a half-life of 360 ns WHY Cu-63 AND Cu-65 WITH S = -3/2 ARE STABLE NUCLIDES For understanding the stable structure of Cu-63 and Cu-65 you may read my STRUCTURE OF Cu-63 AND Cu-65 . The stable structure of these nuclides is based on the symmetrical structure of the Cu-59 with S = -3/2 . In this case the one extra neutron of negative spin of the same horizontal plane fills a symmetrical blank position with respect to the p29n29. While the unstable structure of all nuclides based on the structure of Cu-58 is due to the fact that the deuteron p29n29 of the +HP1 brakes the symmetry. In the following diagram of Cu-58 with 29 protons and 29 neutrons you see that the p29n29 brakes the symmetry of S = -3/2 giving S = +1. ' ' ' DIAGRAM OF Cu-58 WITH S = +1' The core of this structure is the parallelepiped of Mg-24 having six horizontal planes of opposite spins like the +HP1, -HP2, +Hp3, -HP4, +HP5, and -HP6 . Here the 28 deuterons (from p1n1 to p28n28) give a structure of high symmetry with S =0 . Note that the 8 symmetrical deuterons ( from p13n13 to p20n20 with S = 0 at -HP2, +HP3, -HP4, and +HP5) are not shown because they exist in front of Mg-24 and behind it. However the n29p29 with S = +1 of +HP1 gives the total spin S = +1 of Cu-54 and brakes the symmetry.' ' ' p12.......n12' ' -HP6 n11.......p11 ' ' p24........n10.......p10……...n28' ' +HP5 n24………p9.........n9 ……..p28 ' ' n23........p8.........n8............p27' ' -HP4 p23..........n7.........p7............n27 ' ' p22........n6.........p6............n26' ' +HP3 n22………p5.........n5……….p26 ' ' n21….....p4..........n4……...p25' ' -HP2 p21……...n3……..p3……….n25 ' ' p29........n2……....p2' ' +HP1 n29………p1..........n1 ' ' ' NUCLEAR STRUCTURE OF Cu-62, Cu-64, Cu-66, Cu-68 Cu-72, AND Cu-74 WITH S = +1 ''' After a careful analysis I found that the structure of the above unstable nuclides is based on the structure of Cu-58 with S = +1. For example the Cu-74 with S = +1 has 16 extra neutrons of opposite spins. '''NUCLEAR STRUCTURE OF Cu-56, Cu-54, AND Cu-52 In the absence of neutrons we see that the structure of the above unstable nuclides is based not on the Cu-58 with S = +1 but on another structure of Cu-58 with S = +3. In this case the deuteron n11p11 changes the spin from S = -1 to S = +1, because it goes from -HP6 to +HP1 to make horizontal bonds with the n1p2. This change gives S = +2. So adding the S = +1 of the p29n29 we get the total spin S = +2 +1 = +3. That is at +HP1 we have the symmetrical deuterons of n29p29 and p11n11. Nevertheless this new structure brakes again the symmetry because we have not the two symmetrical squares of +HP1 and -HP6 but 4 deuterons with S = +4 of +HP1, and one deuteron with S = -1 at -HP6. Under this new structure of Cu-58 we see that in the structure of Cu-56 with S = +4 we have two absent neutrons with negative spins. That is the total spin is given by S = +3 - 2(-1/2) = +4 Moreover in the Cu-54 or Cu-52 with S = +3 we have analogous absent neutrons of opposite spins. NUCLEAR STRUCTURE OF Cu- 9, Cu-61, Cu-63, Cu-65, Cu-67, Cu-69 Cu-71, ' '''Cu-73, Cu-75, Cu-77, AND Cu-79 WITH S = - 3/2 ' After a careful analysis I found that the structure of the above nuclides is based on the symmetrical structure of Cu-59 with S = -3/2. In this case the p29n29 is at -HP6 with S =-1 having the one extra neutron, say the n30(-1/2), at the same -HP6 and at a symmetrical position. Under this symmetry the Cu-61 receives two more extra neutrons of opposite spins with two bonds per neutron. Since we have only two neutrons we see that their bonds cannot give enough energies to all pn bonds for overcoming the pp and nn repulsions. However in the structure of Cu-63 the 4 more extra neutrons of opposite spins can make two bonds per neutron able to overcome the repulsions. Also the Cu-65 of 6 more extra neutrons with two bonds per neutron lead to the stability. However the Cu-67 is an unstable nuclide because in the two more extra neutrons than those of Cu-65 there is one neutron which makes a single bond leading to the decay. Note that the protons of the two squares of +HP1 and -HP6 can receive 8 neutrons of opposite spins able to make two bonds per neutron. In the case o Cu-67 because of the p29n29 the protons can receive only 7 neutrons with two bonds. Thus, in this structure of Cu-67 there is one neutron which makes single horizontal bond leading to the decay. Similarly in all heavier nuclides the more extra neutrons of opposite spins make single bonds. For example the Cu-79 with S = -3/2 , based on Cu-65 with S = -3/2, has 14 more extra neutrons of opposite spins than those of Cu-65 making single bonds. '''NUCLEAR STRUCTURE OF Cu-57, Cu-55, AND Cu-53 WITH S = -3/2 In the absence of neutrons we see that the structure of the above unstable nuclides is based on the structure of Cu-59 with S = -3/2. For example in the structure of Cu-53 with S =-3/2 we have 6 absent neutrons of opposite spins. ' ' ' ' Category:Fundamental physics concepts